如何将json POST数据作为对象传递给Web API方法？

Question

控制器：

public class CustomersController : ApiController {

  public object Post([FromBody] Customer customer)
        {
            return Request.CreateResponse(HttpStatusCode.OK,
            new
            {
                customer = customer
            });
        }
    }
}

public class Customer
    {
        public string company_name { get; set; }
        public string contact_name { get; set; }
     }

请求：

POST http://localhost:52216/api/customers HTTP/1.1
Accept: application/json, text/javascript, */*; q=0.01
X-Requested-With: XMLHttpRequest
Content-Type: application/x-www-form-urlencoded; charset=UTF-8

{"contact_name":"sdfsd","company_name":"ssssd"}

Answer 1

编辑：2017年10月31日

同样的代码/方法也适用于ASP.NET Core2.0。主要的区别在于，在ASP.NET内核中，将web api控制器和Mvc控制器合并到一个控制器模型中。因此返回类型可能是IActionResult或其实现之一（例如:OKObjectResult)

使用

contentType:"application/json"

并且模型绑定器将json数据绑定到您的类对象。

下面的代码可以正常工作（经过测试）

$(function () {
    var customer = {contact_name :"Scott",company_name:"HP"};
    $.ajax({
        type: "POST",
        data :JSON.stringify(customer),
        url: "api/Customer",
        contentType: "application/json"
    });
});

结果

public class CreateUserViewModel
{
   public int Id {set;get;}
   public string Name {set;get;}  
   public List<TagViewModel> Tags {set;get;}
}
public class TagViewModel
{
  public int Id {set;get;}
  public string Code {set;get;}
}

public class ProductController : Controller
{
    [HttpPost]
    public CreateUserViewModel Save([FromBody] CreateUserViewModel m)
    {
        // I am just returning the posted model as it is. 
        // You may do other stuff and return different response.
        // Ex : missileService.LaunchMissile(m);
        return m;
    }
}

//Build an object which matches the structure of our view model class
var model = {
    Name: "Shyju",
    Id: 123,
    Tags: [{ Id: 12, Code: "C" }, { Id: 33, Code: "Swift" }]
};

$.ajax({
    type: "POST",
    data: JSON.stringify(model),
    url: "../product/save",
    contentType: "application/json"
}).done(function(res) {       
    console.log('res', res);
    // Do something with the result :)
});

如果不使用[FromBody]属性修饰web api方法参数

[HttpPost]
public CreateUserViewModel Save(CreateUserViewModel m)
{
    return m;
}

并发送模型（原始javascript对象，不是JSON格式），而不指定contentType属性值

$.ajax({
    type: "POST",
    data: model,
    url: "../product/save"
}).done(function (res) {
     console.log('res', res);
});

模型绑定将适用于模型上的平面属性，而不是类型复杂的属性/另一种类型。在我们的示例中，id和name属性将正确绑定到参数m，但tags属性将是一个空列表。

$.post("../product/save", model, function (res) {
    //res contains the markup returned by the partial view
    console.log('res', res);
});